Question: We know that $\dfrac{\sqrt{n}}{n-3}>\dfrac{\sqrt{n}}{n}=\dfrac{1}{\sqrt{n}}>0$ for any $n\ge 4$. Considering this fact, what does the direct comparison test say about $\sum\limits_{n=4}^{\infty }\dfrac{\sqrt{n}}{n-3}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A The series converges. (Choice B) B The series diverges. (Choice C) C The test is inconclusive.
Explanation: $\sum_{n=26}^{\infty }{\frac{1}{\sqrt{n}}}$ is a $p$ -series with $p=\dfrac{1}{2}$, so it diverges. Our given series is term-by-term greater than a divergent series, so it also diverges.